chapter 2 fluid statics - test bank and solution manual...

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Chapter 2 Fluid Statics 2.1

Since and , we have and

Let and

2.2 a) b) c) d) e) 2.3 a) b) c) d) e) 2.4 (C)

2.5

2.6 a) b) c) 2.7 a) b) c)


2.8 2.9 (D) 2.10

If no wind is present this would produce a small infiltration since the higher pressure outside would force outside air into the bottom region (through cracks).

2.11 where From the given information since and

By definition where Then Integrate:

Note: we could have used an average , so that

2.12

2.13

The density variation can be ignored over heights of 300 m or less.


2.14

This change is very small and can most often be ignored.

2.15 Eq. 1.5.11 gives But, Therefore,

Integrate, using

Now,

Assume

a) For

b) For

c) For

2.16 Use the result of Example 2.2: a) b) c)


2.17 Use Eq. 2.4.8:

a) . b) c) d)

2.18 Use the result of Example 2.2:

2.19 2.20 a) b) 2.21 Referring to Fig. 2.6a, the pressure in the pipe is If , then

a)

b)

c)

d)

2.22 Referring to Fig. 2.6a, the pressure is . Then

a)

b)

2.23 (C) 2.24 See Fig. 2.6b:

2.25


2.26 (Use ) 2.27 (Use ) 2.28 2.29 2.30 With 2.31 a) or b) or 2.32 or 2.33 (A) 2.34

2.35

Note: In our solutions we usually retain 3 significant digits in the answers (if a number starts with “1” then 4 digits are retained. In most problems a material property is used, i.e., . This is only 3 sig. digits! only 3 are usually retained in the answer!


2.36 Before pressure is applied the air column on the right is 1.2m high. After pressure is

applied, it is m high. For an isothermal process using absolute pressures. Thus,

From a pressure balance on the manometer (pressures in Pa):

or 2.37 2.38

2.39 2.40 a) b) 2.41 2.42 a) b) c)


d) 2.43 For saturated ground, the force on the bottom tending to life the vault is The weight of the vault is approximately The vault will tend to rise out of the ground. 2.44 Find in Table B.5 in the Appendix 2.45 a) where the height if the triangle is b) c) 2.46 a)

b) The centroid is the center of pressure. c)

2.47 (B) The force acts 1/3 the distance from the hinge to the water line: 2.48 a)

b)


c)

d)

2.49

2.50

2.51


2.52 is measured from the surface

From the bottom,

Note: this result is independent of the angle , so it is true for a vertical area or a

sloped area. 2.53

a) b) c) 2.54 Use 2 forces:

2.55 To open, the resultant force must be just above the hinge, i.e., must be just less than .

Let the condition when the gate is about to open:

a) b) c) 2.56 The gate is about to open when the center of pressure is at the hinge.

a)

b)

c)


2.57 (A) The gate opens when the center of pressure is at the hinge:

This can be solved by trial-and-error, or we can simply substitute one of the answers into the equation and check to see if it is correct. This yields

. 2.58

2.59 The dam will topple if the moment about “O” of and exceeds

the restoring moment of and . a)

b)

c) Since it will topple for , it certainly will topple if


2.60 The dam will topple if there is a net clockwise moment about “O” a)

b)

c) Since it will topple for , it will also topple for

2.61

Note: This calculation is simpler than that of Example 2.7. Actually, we could have moved the horizontal force and the vertical force (equal to ) simultaneously to the center of the circle and then This was outlined at the end of Example 2.7

2.62 Since all infinitesimal pressure forces pass thru the center, we can place the resultant forces at the

center. Since the vertical components pass thru the bottom point, they produce no moment about that point. Hence, consider only horizontal forces:

2.63 Place the resultant force at the center of the circular arc. passes thru the hinge

showing that


2.64 (D) Place the force at the center of the circular arc. passes thru the hinge: 2.65 Place the resultant at the circular arc center. passes thru the hinge so that

. Use the water that could be contained above the gate; it produces the same pressure distribution and hence the same

2.66 Place the resultant at center. passes thru the hinge 2.67 The incremental pressure forces on the circular quarter arc pass through the hinge so that

no moment is produced by such forces. Moments about the hinge gives: 2.68 The resultant of the unknown liquid acts thru the center of the circular arc.

passes thru the hinge. Thus we use only . Assume 1 m wide.

2.69 The force of the water is only vertical acting thru the center. The force of the oil can

also be positioned at the center: 2.70 The pressure in the dome is a) The force is

b) From a free-body diagram of the dome filled with oil: Using the pressure from part (a):


2.71 A free=body diagram of the gate and water is shown

2.72 (A)

2.73 W = weight of displaced water. a) b) 2.74 2.75 2.76 2.77 (See Fig. 2.11 c.) 2.78 The forces acting on the balloon are its weight, , the buoyant force , and the weight of

the air in the balloon . Sum forces:


2.79 The forces acting on the blimp are the payload , the weight of the blimp , the buoyant

force , and the weight of the helium :

Of course equipment and other niceties such as gyms, pools, restaurants, etc., would add

significant weight. 2.80 Neglect the buoyant force of air. A force balance yields Specific wt:

2.81 From a force balance a) The buoyant force is found as follows

The that makes the above 2 ’s equal is found by trial-and-error: ? ? 7997 ?

b) Assume m and use the above equations with cm

? 7396

c) Assume m. With cm Trial-and-error for ? ? 7380 ?


2.82 a)

b)

c)

2.83

a) b)

2.84 a)

b) m

c)

2.85 With ends horizontal The displaced volume is since . The depth the cylinder will sink is

This gives (divide by and multiply by : Consequently,


2.86

If the cube is neutral and .

The cube is unstable if Note: Try and to see if . This indicates stability.

2.87 As shown,

2.88

must be directly under

2.89 The centroid is m below the water surface.

Using the Eq. 2.4.47:

The barge is stable

.

2.90

Using the Eq. 2.4.47:


2.91

2.92 a)

b) 2.93 The air volume is the same before and after.

a) b) c) Air fills the space to the dotted line. 2.94 Use Eq. 2.5.2: Assume an air-water surface as shown in the above figure.

a)

b)

c)


2.95 a) Use Eq. 2.5.2. with the peep hole as

position 1. The x axis is horizontal passing thru A. We have

b) 2.96 a) The pressure on the end AB ( is zero at ) is, using Eq. 2.5.2,

b) The pressure on the bottom BC is

c) On the top where position 1 is on the top surface:

2.97 a) The pressure at is 58.29 kPa. At it is

Since the pressure varies linearly over , we

can use an average pressure times the area: b) c)

2.98 Use Eq. 2.5.2. with position 1 at the open end: a) b)


c) 2.99 Use Eq. 2.6.4 with position 1 at the open end: a) b) 2.100 Use Eq. 2.6.4 with position 1 at the open end. a) b) 2.101 Use Eq. 2.6.4 with position 1 at the open end and position 2 at the origin. Given: a) b)


2.102 The air volume before and after is equal a) Using Eq. 2.6.5: b) c) For , part of the bottom is bared. Using Eq. 2.6.5:

Also, d) Following part

(c): 2.103 The answers to Problem 2.102 are increased by 25 000 Pa. a) b) c) d) 2.104

a)

(We used


b)

c)

d)

chapter 2 fluid statics - test bank and solution manual...

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